B 猜数

题目地址：

https://ac.nowcoder.com/acm/contest/5389/B

基本思路：

这题可以直接贪心从小到大改为9就行了，比较简单。
这里我们考虑怎样进一步优化时间复杂度，由于只有0 ~ 9几个数，我们考虑用个桶维护一下每种增加量的出现次数，然后对于每个增加量从9到0数学方法计算最大贪心就好了。

参考代码

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
using namespace std;
#define IO std::ios::sync_with_stdio(false)
#define ll long long
#define rep(i, l, r) for (int i = l; i <= r; i++)
#define per(i, l, r) for (int i = l; i >= r; i--)
#define mset(s, _) memset(s, _, sizeof(s))
#define pb push_back
#define pii pair <int, int>
#define mp(a, b) make_pair(a, b)
#define INF 0x3f3f3f3f

inline int read() {
  int x = 0, neg = 1; char op = getchar();
  while (!isdigit(op)) { if (op == '-') neg = -1; op = getchar(); }
  while (isdigit(op)) { x = 10 * x + op - '0'; op = getchar(); }
  return neg * x;
}
inline void print(int x) {
  if (x < 0) { putchar('-'); x = -x; }
  if (x >= 10) print(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 1e6 + 10;
int n,m,a[maxn];
int memo[10];
signed main() {
  IO;
  n = read(); m = read();
  int sum = 0;
  rep(i,1,n){
    a[i] = read();
    memo[9 - a[i]]++;
    sum += a[i];
  }
  int cnt = 0;
  per(i,9,0){
    if(sum >= m) break;
    int all = memo[i] * i;
    if(sum + all < m) sum += all,cnt += memo[i];
    else{
      int num = ceil((double)(m - sum)/(double)i);
      cnt += num;
      sum += num * i;
    }
  }
  cout << cnt << '\n';
  return 0;
}